# Calculate Force Applied to Shelf from Block of Wood | 500 lbs

• physical?eer
In summary, if the force is applied directly over the supporting brace the screw will likely hold. If the force is applied two feet away from the screw, the torque will likely be several times the 500lb load, so the screw will pull out.f

#### physical?eer

Hello,

I am trying to support a shelf with a block of wood. The block of wood is only 3 inches deep and the shelf 2 feet deep. If the block of wood is capable of supporting 500 lbs, how much weight can the shelf support if the force is applied downward at the edge of the shelf?

This is a real world problem not a homework problem, would appreciate help in calculating this (i.e. what formula should I use).

Last edited by a moderator:
Impossible to say without a diagram showing exactly what you are talking about

russ_watters
Impossible to say without a diagram showing exactly what you are talking about
Black = Wall
The red block of wood is 3 inches deep
The blue shelf is 2 feet deep

If the block of wood is capable of supporting 500 lbs, how much weight can the shelf support if the force is applied downward at the edge of the shelf?

#### Attachments

• Untitled-1.jpg
7.3 KB · Views: 337
Still not anywhere near enough information. Where and how is the triangle attached to the wall? How are the two pieces of wood attached to each other? You say that the shelf can support 500 lbs but you don't say how much torque can the wall studs can support. Even if you assume they can support whatever is needed, the other information is still required.

I'm not trying to give you a hard time here but you seem to have no concept to how things work mechanically and since you neglected to say anything about yourself in your profile, I can't even guess what your level of education is.

Still not anywhere near enough information. Where and how is the triangle attached to the wall? How are the two pieces of wood attached to each other? You say that the shelf can support 500 lbs but you don't say how much torque can the wall studs can support. Even if you assume they can support whatever is needed, the other information is still required.

I'm not trying to give you a hard time here but you seem to have no concept to how things work mechanically.

The block of wood is in the middle of the shelf. The shelf is 3 feet wide and 2 feet deep, it is supported in the middle of the shelf. I'm going from the block of wood into the studs with two screws at a 90* angle. From what I understand each screw is capable of supporting >500lbs. I have about 1.5" of the screw going into the stud; which is a standard #8 screw. The wall screws will give out before the studs will.

I was trying to simplify this equation, so if you'd like, assume that the shelf is only as wide as the block of wood (i.e. no x-axis).

I hope that helps, and thank you.

The block of wood is in the middle of the shelf. The shelf is 3 feet wide and 2 feet deep, it is supported in the middle of the shelf. I'm going from the block of wood into the studs with two screws at a 90* angle. From what I understand each screw is capable of supporting >500lbs. I have about 1.5" of the screw going into the stud; which is a standard #8 screw. The wall screws will give out before the studs will.

I was trying to simplify this equation, so if you'd like, assume that the shelf is only as wide as the block of wood (i.e. no x-axis).

I hope that helps, and thank you.
OK, that helps. I think basically you are saying that all joints should be taken as capable of supporting whatever load is applied. That's not unreasonable although your statement that "each screw is capable of supporting >500 lbs is not as helpful as you seem to think it is since what it means is that if the force were applied directly in parallel with the screw, which is not quite what we have here, although it's close if you put the top screw as high up as possible in the triangle. The bottom screw will really only be used for shear stress so it's holding capacity is irrelevant unless it is just below the top screw , so they entire load is supported by one screw that can hold something that is an unspecified amount more than the expected load. With the force applied two feet away from the screw, the torque will likely (I haven't done any computation) be several times the 500lb load, so the screw will pull out.

THEN you totally change the problem by going from a force applied two feet out to a force applied directly over the supporting brace. In that case it will probably hold, although if you really only have one brace under the middle of the board, any lateral imbalance in the load will cause a whole 'nother set of problems.

"each screw is capable of supporting >500 lbs is not as helpful as you seem to think it is since what it means is that if the force were applied directly in parallel with the screw.

I'm not looking for exact numbers so what if we overly simplified this problem as follows (see image). Does that help?

#### Attachments

• 1.png
16.9 KB · Views: 684
I'm not looking for exact numbers so what if we overly simplified this problem as follows (see image). Does that help?View attachment 231431
Yes, that's a simple one. With no further support other than the glue at the back of the plank, and assuming a 3/4" plank, the torque on the glue joint will be at least several thousand pounds and the joint will fail.

I know you are trying to simplify things to get at an answer but your obvious lack of understanding of simple mechanics is getting in the way of your producing a meaningful question. I don't think I can be of help to you so I'll bow out of this thread and leave it to some of our mechanical engineers.

berkeman
If anyone has a formula to determine the force created when applied at a distance I'd appreciate it. I'm surprised this is turning into such a complicated question.

Trick is, there are several forces acting on the support in different directions and you are asking for one formula describing them all.

Trick is, there are several forces acting on the support in different directions and you are asking for one formula describing them all.
And doing so without adequately describing ANY of them.

From what I understand each screw is capable of supporting >500lbs. I have about 1.5" of the screw going into the stud; which is a standard #8 screw. The wall screws will give out before the studs will.
A #8 wood screw into a stud is capable of supporting 500 pounds? Where did you get this from? And in tension, or in shear, or in some combination?

A #8 wood screw into a stud is capable of supporting 500 pounds? Where did you get this from? And in tension, or in shear, or in some combination?
Grain (across, with, OSB, laminated)? Species? Cured/seasoned? "Manufactured Forest Products?" There are more missing specifications than there are furnished; 500 pound loads are significant in this application, shelving.

berkeman
OP seems unfamiliar with free body diagrams.
Maybe we need to show him how to calculate the tension and shear loads on the upper screw ? Sum the torques ?

That way he might understand better why we're reluctant to say "Build it this way".

He can tinker with a spreadsheet and see the effect of changing height of that 3 inch block.and location of the screws.
Myself i'd make a triangular support almost out to the edge of the shelf. I use oak about 3/4 inch thick. For heavy loads i dowel the joints.

Here is a step by step for using Douglas Fir for the wall stud and the block.
It is NOT COMPLETE, the shelf board may crack at the front of the block, the screw head may pull thru the board, and the 45° corners of the triangular block may break off.

Pull-out resistance (p) from:
https://www.fpl.fs.fed.us/documnts/fplgtr/fplgtr190/chapter_08.pdf
pgs 10,11

This formula is good for up to 2 inches of thread engagement of a #8 screw.

p = 15,700G^2DL
G= specific gravity of wood
D= shank dia. of screw
L= thread engagement on wood

#8 wood screw: shank dia.D = 0.164
Wood: Sp.Gravity of Douglas Fir G= 0.5

p = 15700xGxGx0.164
p = 643 lbs per inch of embedded thread length

Assuming the screw holding board to block is 5/8 inch from back edge, the distance from screw to front of block is 2.375in.

The front edge of the board to the pivot point at front of block is 21in.

This gives a lever action, or ratio, of 21/2.375 = 8.84. So each pound of force at the front of the board creates 8.84 pounds trying to pull the screw out.

The screw can hold 640 pounds per inch, which translates to 640/8.84 = 72 pounds at the front of board, per inch of thread engagement.

This same calculation can be used to approximate the forces between the wall and the block. Note also that the bottom edge of the block will exert about the same force on the wall material as the force pulling on the screw. This may crush the wall material, especially if it is plaster or plasterboard.

I'll leave the fine points to the mechanical experts here.

Cheers,
Tom

jim hardy
@ physical?eer
@physical?eer @ worked fourth time

Annotating your original drawing,

'reaction' will try to pull the screw out of the wall.
If the screw has pullout of 500 lbs that's your max allowable 'reaction' without any safety factor..
Force = Reaction/8 = 500/8 = 62,5 pounds is your max load without any safety factor
that's why a taller block is better.
The vertical load, shear, on the screw will be 'Force' plus weight of the shelf parts.

Work that example until it makes sense . You'll begin developing a 'feel' for leverage and mechanics.

If you were trying to tell us the block can support 500 pounds of vertical load, well, that means i answered the wrong question.

old jim

#### Attachments

• shelfbracket.jpg
15.9 KB · Views: 442
Last edited:
Bystander